Class 12th

$l^{-}+H_2{O}_2+2H^{+}\to l_2+2H_{2}O$

(-1) oxidation (0)

Here, I is reducing agent

${}_{}$ $\therefore H_2{O}_2$  behaves like oxidizing agent

In Ce & Eu (+3) oxidation state is more stable

$\therefore C{e}^{+4}+e^{-}\to C{e}^{+3}\left(Reduction\right)$ so CeO₂ is oxidising agent

$E{u}^{+2}\to E{u}^{+3}+e-\left(oxidation\right)$ so EuSO₄ is reducing agent

Given $l_{m,n}={\displaystyle {\int }_0^1{x}^{m-1}{\left(1-x\right)}^{n-1}dx............\left(i\right)}$  

put 1 - x = $t\{\begin{array}{l}x=0,t=1\\ x=1,t=0\end{array}$  

dx = -dt

From (i) $l_{m,n}={\displaystyle {\int }_1^0{\left(1-t\right)}^{m-1}.t^{n-1}\left(-dt\right)}$  

$l_{m,n}={\displaystyle {\int }_0^1{t}^{n-1}{\left(1-t\right)}^{m-1}dt={\displaystyle {\int }_0^1{x}^{n-1}{\left(1-x\right)}^{m-1}dx..........\left(ii\right)}}$                              

(i)   $l_{m,n}={\displaystyle {\int }_{\infty }^0\frac{1}{{\left(1+y\right)}^{m-1}}}{\left(1-\frac{1}{1+y}\right)}^{n-1}\left(\frac{-dy}{{\left(1+y\right)}^2}\right)={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}}{{\left(1+y\right)}^{m+n}}dy}..........\left(iii\right)$

Similarly by (ii) $l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{m-1}}{{\left(y+1\right)}^{m+n}}dy..........\left(iv\right)}$  

Adding (iii) & (iv) $2l_{m,n}={\displaystyle {\int }_0^{\infty }\frac{y^{n-1}+y^{m+1}}{{\left(y+1\right)}^{m+n}}}$  

Putting   $\frac{1}{z}\{\begin{array}{l}y=1,z=1\\ y=\infty ,z=0\end{array}$ $\Rightarrow dy=\frac{-1}{z^2}dz$  

Hence  $l_{m,n}={\displaystyle {\int }_0^1\frac{x^{m-1}+x^{n-1}}{{\left(1+x\right)}^{m+n}}}$ dx = a lm, n

-> a = 1

Given $f\left(x\right)=2x^5+5x^4+10x^3+10x^2+10x+10$

$f\left(-1\right)=3>0\&f\left(-2\right)=-34<0$           

So at least one root will lie in (-2, -1)

now $f\text{'}\left(x\right)=10x^4+20x^3+30x^2+20x+10$  

$=10\left[x^4+2x^3+3x^2+2x+1\right]$            

$=10x^2{\left(x+\frac{1}{x}+1\right)}^2>0\forall x\in R$           

So, f(x) be purely increasing function so exactly one root of f(x) that will lie in (-2, 1). Hence |a| = 2

Kindly consider the following figure

Given $P_n={\alpha }^n+{\beta }^n,P_{n-1}=11\&P_{n+1}=29$  

$P_n={\alpha }^{n-2}.{\alpha }^2+{\beta }^{n-2}.{\beta }^2.........\left(i\right)$                             

Now quadratic equation having roots a & b will be x² – (a + b)x + ab = 0

i.e.         x² – x – 1 = 0,     put x = a and put x = b

So          a² = a + 1           & b² = b + 1

(i)  $P_n={\alpha }^{n-2}\left(\alpha +1\right)+{\beta }^{n-2}\left(\beta +1\right)$    

$P_n=P_{n-1}+P_{n-2}$                        

-> ${P_n}^2=234$

Kindly consider the following figure

Glycosidic linkage is present in lactose between C₁ of galactose & C₄ of glucose.

Given curves $\frac{x^2}{9}+\frac{y^4}{4}=1.........\left(i\right)$  

&       $x^2+y^2=\frac{31}{4}...........\left(ii\right)$   

Equation of any tangent to (i) be y = mx +    $\sqrt[]{9m^2+4}............\left(iii\right)$

For common tangent (iii) also should be tangent to (ii) so by condition of common tangency

$9m^2+4=\frac{31}{4}\left(1+m^2\right)$          

OR 36m² + 16 = 31 + 31m²

->m² = 3